In this paper, we are concerned with the existence of positive solutions for boundary value problems of nonlinear fourth-order differential equations
u(4)+c(x)u=λa(x)f(u),x∈(0,1),u(0)=u(1)=u″(0)=u″(1)=0,
where a(x) may change signs. The proof of main results is based on Leray-Schauder's fixed point theorem and the properties of Green's function of the fourth-order differential operator Lcu=u(4)+c(x)u.
Citation: Yanhong Zhang, Li Chen. Positive solution for a class of nonlinear fourth-order boundary value problem[J]. AIMS Mathematics, 2023, 8(1): 1014-1021. doi: 10.3934/math.2023049
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In this paper, we are concerned with the existence of positive solutions for boundary value problems of nonlinear fourth-order differential equations
u(4)+c(x)u=λa(x)f(u),x∈(0,1),u(0)=u(1)=u″(0)=u″(1)=0,
where a(x) may change signs. The proof of main results is based on Leray-Schauder's fixed point theorem and the properties of Green's function of the fourth-order differential operator Lcu=u(4)+c(x)u.
Nonlinear mathematical models [1,2] were widely used in many fields. In particular, boundary value problems of nonlinear differential equations have received extensive attention and have been intensively studied in the past thirty years, see [3,4]. We point out that boundary value problems for second order differential equations, see, for example [5,6,7,8,9] and the references therein. While studies about boundary value problems of nonlinear fourth-order differential equations are much more less. One of the earliest papers about boundary value problems of nonlinear fourth-order differential operator is [10] from R. Ma and H. Wang, there they concerned the following problem
y(4)−h(x)f(y(x))=0,x∈(0,1) |
with boundary condition
y(0)=y(1)=y″(0)=y″(1)=0 |
or
y(0)=y′(1)=y″(0)=y‴(1)=0. |
By the fixed point theorem in cone, they proved the existence of positive solutions under the conditions that f is either superlinear or sublinear. In another paper [11], the author obtained the positive solution of the following problem
u(4)+βu″−αu=f(t,u),x∈(0,1)u(0)=u(1)=u″(0)=u″(1)=0 |
by the fixed point theorem in cone. R.Vrabel [12] studied the upper solution and lower solution of the problem
y(4)(x)+λy″(x)=h(x,y(x)),x∈(0,1)y(0)=y(1)=y″(0)=y″(1)=0. |
There are many other papers we will not list but we find that they have a common point, that is, the fourth-order differential operators they dealt with can be resolved into composition of two second-order positive linear operators. And therefore, the corresponding Green's function for fourth-order linear operator is the form of the product of two Green's functions for second-order linear operators.
In a recently paper [13], Drábet discussed the existence of positive solutions for the following fourth-order linear problem
u(4)+c(x)u=h(x),x∈(0,1),u(0)=u(1)=u″(0)=u″(1)=0. |
Obviously, the fourth-order differential operator can not be resolution into composition of two second-order positive linear operators. For more results on nonlinear fourth-order differential operator problems we can refer to [14,15].
Based on the above literature inspiration. We now consider the fourth-order nonlinear equation with Dirichlet boundary conditions
u(4)+c(x)u=λa(x)f(u), | (1.1) |
u(0)=u(1)=u″(0)=u″(1)=0, | (1.2) |
where c(x),a(x) satisfy some conditions that we will give bellow, especially, a(x) may change signs.
We make the following assumptions throughout the paper:
(A1) −π4<c(x)<c0,
(A2) f:R+→R is continuous, and f(0)>0,
(A3) a:[0,1]→R is continuous with a(x)≢0, and there exists a constant K>0 such that
∫10G(x,y)a+(y)dy≥K∫10G(x,y)a−(y)dy |
for every x∈(0,1), where a+(resp. a−) is the positive (resp. negative) part of a, c0 is the constant given in [13], and G(x,y) is the Green's function of Lc with boundary conditions (1.2).
Our main result is as follows:
Theorem 1.1. Let (A1)–(A3) hold. Then there exists a positive number λ∗ such that (1.1) and (1.2) have a positive solution for 0<λ<λ∗.
Remark 1.1. Since the fourth-order differential operator can not be resolution into composition of two second-order positive linear operators, as a result, the Green's function have no explicit expression. So the method or technic used in [10,11,12] does not work. To deal with the new case and the difficult it brings, we are inspired by the method to second-order elliptic boundary value problems in [8], and the result that the fourth-order operator u(4)+c(x)u is strictly inverse positive in [13,16]. Thanks to the existence and its properties of the Green's function given in [17,18,19], we obtain the existence of a positive solution to the problems (1.1) and (1.2).
In this section we present two important lemmas. The main method we use is the fixed point theorem of Leray-Schauder type. We refer interested readers to the literature [20,21].
Set
W={u∈C4([0,1]):u(0)=u(1)=u″(0)=u″(1)=0}, |
and let the linear operator Lc:W→C([0,1]) defined by
Lcu=u(4)+c(x)u. |
Then the boundary value problems (1.1) and (1.2) are equivalent to the operator equation
Lcu=λa(x)f(u). |
Lemma 2.1. Let (A1) hold. Then Lc is strictly inverse positive, and therefore it has a positive Green's function.
Proof. Lc is strictly inverse positive, we can refer to [13,16] and the reference therein. From the definition of Lc is strictly inverse positive there and the well-known truth that
Lcu=h(x) |
is equivalent to
u(x)=∫10G(x,y)h(y))dy, |
we can get the positiveness of the Green's function G(x,y) immediately.
Lemma 2.2. Let (A1)–(A3) hold, and let 0<δ<1. Then there exists a positive number ˉλ such that, for 0<λ<ˉλ, the problem
u(4)+c(x)u=λa(x)+f(u) | (2.1) |
u(0)=u(1)=u″(0)=u″(1)=0 | (2.2) |
has a positive solution ˜uλ with |˜uλ|→0 as λ→0, and
˜uλ(x)≥λδf(0)p(x),x∈(0,1), |
where p(x)=∫10G(x,y)a+(y)dy.
Proof. It follows from Lemma 2.1 that Lc is strictly inverse positive, and therefore it has a positive Green's function G(x,y). For each u∈C([0,1]), let
Au(x)=λ∫10G(x,y)a+(y)f(u(y))dy,x∈[0,1]. |
Then the fixed points of A are solutions of problems (2.1) and (2.2). We now verify the condition of Leray-Schauder fixed point theorem to show that A has a fixed point for λ small.
Firstly, A:C([0,1])→C([0,1]) is completely continuous by the assumptions and Arzela-Ascoli theorem.
Secondly, we find a bounded open set Ω with 0∈Ω in C([0,1]), such that for u∈C(ˉΩ) and θ∈(0,1), if u=θAu, then uˉ∈∂Ω.
By (A2), the function g(s)=f(s)f(0) is continuous and g(0)=1, since 0<δ<1, we can choose ε>0 such that
f(s)>δf(0)s∈[0,ε]. |
Also we have
|Au|0≤λ|p|0˜f(|u|0)≤λ|p|0˜f(ε),u∈[0,ε], |
where ˜f(t)=max0≤s≤tf(s), and |⋅|0 is the usual norm in C([0,1]).
Suppose λ<12|p|0˜f(ε)=:ˉλ, then there exists a Aλ∈(0,ε) such that
˜f(Aλ)Aλ=12λ|p|0. |
Let Ω={u∈C([0,1]):|u|0<Aλ} and θ∈(0,1) such that u=θAu. Then we have
|u|0≤|Au|0≤λ|p|0˜f(|u|0), |
or
˜f(|u|0)|u|0≥1λ|p|0. |
So u≠Aλ, which means uˉ∈∂Ω.
By the Leray-Schauder fixed point theorem, A has a fixed point ~uλ in Ω for 0<λ<ˉλ, that is, problems (2.1) and (2.2) have a positive solution ˜uλ with ˜uλ≤Aλ<ε. Notice that Aλ→0 as λ→0, |˜uλ|→0 as λ→0 and
˜uλ(x)=A˜uλ(x)≥λδf(0)p(x),x∈(0,1). |
The proof is completed.
Proof of Theorem 1.1. Let q(x)=∫10G(x,y)a−(y)dy, recall that p(x)=∫10G(x,y)a+(y)dy. By (A2),
q(x)≤1Kp(x). |
From the proof of Lemma 2.2 that g(0)=1, there is a α∈(0,1) and we can choose 1<σ<K, such that f(s)<σf(0), and γ=σK∈(0,1), then we have
q(x)f(s)≤γf(0)p(x) | (3.1) |
for s∈[0,α],x∈(0,1). Fix δ∈(0,1) and let λ∗>0 be such that
|˜uλ|0+λδf(0)|p|0≤α | (3.2) |
for 0<λ<λ∗, where ˜uλ is the solution of (2.1) and (2.2) given by Lemma 2.2, and
|f(s)−f(t)|≤δ−γ2⋅f(0) | (3.3) |
for s,t∈[−α,α] with |s−t|≤λ∗δf(0)|p|0.
Let 0<λ<λ∗, we look for a solution uλ=˜uλ+vλ. Since ˜uλ is the solution of (2.1) and (2.2), then vλ solves
Lcvλ=λa+[f(˜uλ+vλ)−f(˜uλ)]−λa−f(˜uλ+vλ),x∈(0,1),vλ(0)=vλ(1)=v″λ(0)=v″λ(1)=0. |
For each w∈C([0,1]), let v=Aw be the solution of
Lcv=λa+[f(˜uλ+w)−f(˜uλ)]−λa−f(˜uλ+w),x∈(0,1),v(0)=v(1)=v″(0)=v″(1)=0, |
where the operator A is as in Lemma 2.2, we have
Aw(x)=λ∫10G(x,y)a+(y)[f(˜uλ(y)+w(y))−f(˜uλ(y))]dy−λ∫10G(x,y)a−(y)f(˜uλ(y)+w(y))dy,x∈[0,1], |
and A is completely continuous.
Let
¯Ω′={v∈C([0,1]);|v|0≤λδf(0)|p|0}, |
if v∈C(¯Ω′) and θ∈(0,1), such that v=θAv, that is
v(x)=λθ∫10G(x,y)a+(y)[f(˜uλ(y)+v(y))−f(˜uλ(y))]dy−λθ∫10G(x,y)a−(y)f(˜uλ(y)+v(y))dy,x∈[0,1], |
we are going to show that
|v|0≠λδf(0)|p|0. |
Suppose the contrary that |v|0=λδf(0)|p|0. Then by (3.2) and (3.3), we get
|˜uλ+v|0≤|˜uλ|0+|v|0≤α, |
and
|f(˜uλ+v)−f(˜uλ)|0≤δ−γ2⋅f(0), |
together with (3.1) implies that
|v(x)|≤λ⋅δ−γ2⋅f(0)p(x)+λγf(0)p(x)=λ⋅δ+γ2⋅f(0)p(x),x∈[0,1], | (3.4) |
and
|v|0≤λ⋅δ+γ2⋅f(0)|p|0<λδf(0)|p|0, |
a contradiction.
By the Leray-Schauder fixed point theorem, A has a fixed point vλ in ¯Ω′ with |vλ|0≤λδf(0)|p|0. Hence vλ satisfies (3.4), and using Lemma 2.2, we obtain
uλ(x)=˜uλ(x)+vλ(x)≥˜uλ(x)−|vλ(x)|≥λδf(0)p(x)−λ⋅δ+γ2⋅f(0)p(x)=λ⋅δ−γ2⋅f(0)p(x)>0. |
We have proved that uλ is a positive solution of (1.1) and (1.2).
In this paper, we mainly study the existence of solutions to a class of nonlinear fourth-order Dilrichlet boundary value problems through Leray-Schauder's fixed point theorem, and show the asymptotic behavior of the solution as λ changes. In the future, we can try to construct such solutions, give the properties of the solutions, or study numerical solutions for such problems.
The authors would like to thank the referees and editors for providing very helpful comments and suggestions. This work is supported by Natural Science Foundation of Gansu Province (No. 21JR1RA317) and Young Doctor Fund project of Gansu Province (No. 2022QB-173).
The authors declare no conflicts of interest in this paper.
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